Question

In a hydrogen atom (i.e., one stationary proton and one orbiting electron), the angular velocity of the electron is 10 x 10^6 radians per second. What is the radius of the electron orbit? Assume uniform circular motion.

Answer 1

Assuming the centripetal force is provided by the Coulomb attraction between the electron and the proton we have that,

`F_c = k (e^2)/(R^2)`

Here,

k = Coulomb's Constant

R = Distance

e = Electron charge

And by the centripetal force,

`F_c = m_e R\omega^2`

`m_e` = Mass of electron

R = Radius

`\omega` = Angular velocity

Equation both expression,

`k (e^2)/(R^2) = m_e R\omega^2`

Replacing,

`R^3 = ((9*10^9)(1.6*10^(-19))^2)/((9.11*10^(-31))(10^(6))^2)`

`R^3 = 2.52908*10^(-10) m`

`R = 6.32394*10^(-4)m`

Therefore the radius of the electron orbit is
`6.32394*10^(-4)m`